Bài 14 : 

a. 55/66 = 55 : 11/66 : 11 = 5/6          b. -33/88 = -33 : 11/88 : 11 = -3/8        c. 2020/8080 = 2020 : 2020/8080 : 2020 = 1/4      

d. 2121/-4242 = 2121 : (-101)/(-4242) : (-101) = 21/42 = 1/2     

e. 130130/-260260 = 130130 : (-1001)/(-260260) : ( -1001 ) = -130/260 = -1/2

f. 1616/-1313 = 1616 : (-101)/(-1313) : (-101 ) = -16/13

 Chúc bạn học tốt nha !

Bài 14 khá đơn giản nên mình sẽ bỏ qua, bạn tự làm đc mà xD

15. a) \(\frac{13\cdot9-13\cdot2}{25-12}=\frac{13\left(9-2\right)̸}{13}=7\)

b) \(\frac{42-14\cdot8}{7+21\cdot3}=\frac{14\cdot3-14\cdot8}{7+63}=\frac{14\left(3-8\right)}{70}=\frac{14\cdot\left(-5\right)}{70}=\frac{-70}{70}=-1\)

c) \(\frac{9\cdot6-18\cdot4}{12\cdot3}=\frac{\left(9\cdot2\right)\cdot\left(6:2\right)-18\cdot4}{36}=\frac{18\cdot3-18\cdot4}{36}=\frac{18\left(3-4\right)}{36}=\frac{-18}{36}=\frac{-1}{2}\)

d) \(\frac{5\cdot5^2}{9\cdot10^2-4\cdot10^2}=\frac{5\cdot5^2}{10^2\left(9-4\right)}=\frac{5\cdot5^2}{10^2\cdot5}=\frac{1}{2^2}=\frac{1}{4}\)

e) \(\frac{\left(-7\right)\cdot3-4\cdot\left(-6\right)}{\left(-5\right)\cdot3-2\cdot3}=\frac{-21+24}{-15-6}=\frac{3}{-21}=\frac{-1}{7}\)

f) \(\frac{\left(-17\right)\cdot13+17\cdot2}{-11\cdot2-11\cdot19}=\frac{-\left[17\cdot\left(13+2\right)\right]}{-\left[11\cdot\left(2-19\right)\right]}=\frac{-17\cdot13-\left(-17\right)\cdot2}{-11\cdot2+\left(-11\right)\cdot19}=\frac{-17\left(13-2\right)}{-11\left(2+19\right)}=\frac{-17\cdot11}{-11\cdot19}=\frac{-17}{-19}=\frac{17}{19}\)

* Hơi dài xD , mong bạn thông cảm *

a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)

\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)

b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)

\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)

a ) 7. b ) − 1 c ) − 1 2

d ) 1 4 e ) − 1 7 f ) 17 21

a) \(\dfrac{56}{42}=\dfrac{28}{21}=\dfrac{4}{3}\)

\(\dfrac{16}{24}=\dfrac{2}{3}=\dfrac{8}{12}=\dfrac{40}{60}\)

b) \(\dfrac{40}{25}=\dfrac{8}{5}\)

\(\dfrac{63}{81}=\dfrac{7}{9}\)

\(\dfrac{36}{60}=\dfrac{3}{5}\)

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

a, =5/14

b, = 1/6

c, = 3/2

d, = 8/7

5/14

1/6

3/2

8/7

a,5/14

b,1/6

c,3/2

d,8/7

a) \(\dfrac{22}{55}=\dfrac{2}{5}\)

b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)

c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)

d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)

\(\dfrac{22}{55}=\dfrac{2}{5}\)                                          \(\dfrac{-63}{81}=\dfrac{-7}{9}\)

\(\dfrac{2.14}{7.8}=\dfrac{28}{56}=\dfrac{1}{2}\)                             \(\dfrac{49+7.49}{49}=8\)

a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)

b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)

a) \(\dfrac{3x^2+6xy}{6x^2}=\dfrac{3x\left(x+2y\right)}{6x^2}=\dfrac{x+2y}{2x}\)

b) \(\dfrac{2x^2-x^3}{x^2-4}=\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x^2}{x+2}\)

c) \(=\dfrac{x+1}{x^3+1}=\dfrac{x+1}{\left(x+1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`

`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`

`= -x^2/(x+2)`

`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`